1) lim _(xarrow 0)(xe^x)/(e^x)-1

To evaluate the limit , we can use L'Hôpital's Rule since the expression is in an indeterminate form .First, let's apply L'Hôpital's Rule. L'Hôpital's Rule states that if \(\lim_{x \rightarrow c} \frac{f(x)}{g(x)}\) results in an indeterminate form or , then: provided the limit on the right-hand side exists. Here, \(f(x) = xe^x\) and \(g(x) = e^x - 1\).First, we need to find the derivatives of \(f(x)\) and \(g(x)\): Now, applying L'Hôpital's Rule: We can cancel out in the numerator and the denominator: Now, substitute : Therefore, the limit is: